Question

Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of ((20.0 m/s)i - (4.0 m/s)j). During the 3 ms that the racket and ball are in contact, the net force on the ball is constant and equal to (-(380 N)i + (110 N)j.a) What are the x and y-components of the impulse of the net force applied tothe ball ?b) What are the x and y-components of the final velocity of the ball ?

Answer 1

`\Delta P = -1.14 \hat i + 0.33\hat j`

Part b)

`v_f = 0.03 \hat i + 1.79 \hat j`

Step-by-step explanation:

Part a)

As we know that impulse is due to the force applied for small time due to which momentum is changed

so we will have

`\Delta P = F\Delta t`

so we will have

`F = -380 \hat i + 110 \hat j`

also we know that

`\Delta t = 3 ms`

now impulse is given as

`\Delta P = (-380\hat i + 110 \hat j)(3 * 10^(-3))`

`\Delta P = -1.14 \hat i + 0.33\hat j`

Part b)

As we know that

`\Delta P = m(v_f - v_i)`

`(-1.14 \hat i + 0.33 \hat j) = ((0.560)/(9.81))(v_f - (20\hat i - 4\hat j))`

`-19.97 \hat i + 5.79\hat i + 20 \hat i - 4\hat j = v_f`

`v_f = 0.03 \hat i + 1.79 \hat j`